CF1223F. Stack Exterminable Arrays

Solution

奇怪的套路增加了。

大概就是$[l,r]$能匹配完可以转化为$[1,l-1]$匹配后的状态和$[1,r]$匹配后的状态相同，因此$hash$判断即可。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+9;
const int MAXN=1000005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
ull h[MAXN];
map<ull,int> Map;
int a[MAXN],f[MAXN],stk[MAXN],SZ=1e9+7;
signed main()
{int Case=read();while (Case--){ll ans=0;int n=read(),top=0;for (int i=1;i<=n;i++) a[i]=read(),f[i]=0;Map.clear(),Map[0]=n+1,f[n+1]=0;for (int i=n;i>=1;i--){if (top&&stk[top]==a[i]) top--;else stk[++top]=a[i],h[top]=h[top-1]*SZ+a[i];if (Map[h[top]]) f[i]=f[Map[h[top]]]+1;Map[h[top]]=i;ans+=f[i];}printf("%lld\n",ans);}return 0;
}